Quantum physics and the photoelectric effect

At the start of the 20th century physics discovered the dual nature of light: in interference and diffraction it behaves as a wave, but in emission and absorption it acts as a stream of separate energy packets — quanta (photons). The photoelectric effect is the clearest proof. Solar panels, photocells, digital camera sensors and automatic doors are based on it.

§1. Planck’s hypothesis and the photon

In 1900 Planck, to explain the radiation of hot bodies, proposed that energy is emitted not continuously but in portions. Einstein (1905) went further: light also travels and is absorbed in such portions. This particle is the photon.

• A photon always moves at c; its rest mass is zero.
• In the microworld energy is conveniently measured in electronvolts: 1 eV = 1.6·10⁻¹⁹ J.
• Light pressure is due to photon momentum (Lebedev’s experiment).

Worked example

Energy and momentum of a photon with λ = 500 nm?

1. E = hc/λ = 6.63·10⁻³⁴·3·10⁸/5·10⁻⁷ ≈ 3.98·10⁻¹⁹ J.
2. In electronvolts: 3.98·10⁻¹⁹/1.6·10⁻¹⁹ ≈ 2.5 eV.
3. p = h/λ = 6.63·10⁻³⁴/5·10⁻⁷ ≈ 1.33·10⁻²⁷ kg·m/s.

Worked example

How many photons per second does a 1 mW laser with λ = 663 nm emit?

1. One photon: E = 6.63·10⁻³⁴·3·10⁸/6.63·10⁻⁷ = 3·10⁻¹⁹ J.
2. N = Pt/E = 10⁻³·1/3·10⁻¹⁹ ≈ 3.3·10¹⁵.

Why was the quantum idea needed? Classical theory predicted that a hot body should radiate infinite energy at short wavelengths (the “ultraviolet catastrophe”), which experiment did not show. By assuming that energy is emitted in portions E = hν, Planck obtained a formula that matched experiment perfectly. A quantum is so small that energy seems continuous in everyday life: a 1 W source of visible light emits more than 10¹⁸ photons per second.

§2. The photoelectric effect and its laws

The external photoelectric effect is the emission of electrons from a substance (usually a metal) under light. Hertz observed it (1887), and Stoletov studied it in detail by experiment. The laws:

1. The saturation current (electrons emitted per second) is proportional to the light intensity.
2. The maximum kinetic energy of photoelectrons does not depend on intensity; it increases linearly with the light’s frequency.
3. Each substance has a threshold (red limit): below that frequency (above that wavelength) there is no photoelectric effect at all.
4. The effect is practically instantaneous — it starts as soon as light falls.

Wave theory cannot explain laws 2 and 3: according to it, brighter light should give the electrons more energy.

§3. Einstein’s equation

The photon’s energy goes to a single electron: part of it pulls the electron out of the metal (the work function A), the rest becomes kinetic energy:

The work function depends only on the metal (typically 2–5 eV).

Worked example

Light of λ = 400 nm falls on a metal with a 2 eV work function. Kinetic energy, stopping voltage and threshold wavelength?

1. Photon energy: E = 6.63·10⁻³⁴·3·10⁸/4·10⁻⁷ ≈ 4.97·10⁻¹⁹ J ≈ 3.1 eV.
2. E_k = 3.1 − 2 = 1.1 eV, so U_s = 1.1 V.
3. λ₀ = hc/A = 1.989·10⁻²⁵/(2·1.6·10⁻¹⁹) ≈ 6.2·10⁻⁷ m = 620 nm.

Worked example

Light of frequency 5ν falls on a metal with threshold ν, and light of 4ν on a metal with threshold 2ν. Ratio of kinetic energies?

1. E₁ = 5hν − hν = 4hν, E₂ = 4hν − 2hν = 2hν.
2. E₁/E₂ = 2.

Worked example

Lowest frequency that causes photoemission from a metal with a 2.5 eV work function?

1. A = 2.5·1.6·10⁻¹⁹ = 4·10⁻¹⁹ J.
2. ν₀ = A/h = 4·10⁻¹⁹/6.63·10⁻³⁴ ≈ 6·10¹⁴ Hz.

The graph of E_k against ν is a straight line with slope h that cuts the frequency axis at ν₀ — an experimental confirmation of Einstein’s equation.

§4. Wave–particle duality

Light has both wave and particle properties. De Broglie (1924) proposed that this holds for all particles:

Electron diffraction confirmed the idea; the electron microscope is based on it.

Davisson and Germer (1927) confirmed the wave nature of electrons by observing electron diffraction from a crystal. For everyday objects the wavelength is negligible: a 0.1 kg ball at 10 m/s has λ = h/(mv) = 6.63·10⁻³⁴/1 ≈ 6.6·10⁻³⁴ m, which no instrument could detect, so wave properties show up only in the microworld.

The internal photoelectric effect is the creation of free charge carriers in a semiconductor by light (photoresistors, solar cells).

Worked example

De Broglie wavelength of an electron moving at 10⁶ m/s?

1. λ = 6.63·10⁻³⁴/(9.1·10⁻³¹·10⁶).
2. λ ≈ 7.3·10⁻¹⁰ m ≈ 0.73 nm — the order of X-ray wavelengths.

§5. Uses of the photoelectric effect; light pressure

The external photoelectric effect is used in vacuum photocells: electrons leaving an illuminated cathode carry a current in the circuit. Such cells in automatic doors, metro turnstiles and burglar alarms “notice” when a light beam is blocked. Photomultipliers detect very faint light, down to single photons.

The internal photoelectric effect happens in semiconductors: a photon frees a bound electron and the conductivity rises. Photoresistors, photodiodes, digital camera sensors and solar cells rely on it. In a solar cell a p-n junction separates the freed charges and gives an EMF directly; modern silicon panels are usually 15–22 % efficient.

Light pressure. Photons carry momentum, so light pushes on a surface it strikes. P. N. Lebedev first measured it (1899–1900). If light of intensity I (W/m²) is fully absorbed:

The pressure is tiny, but it explains why comet tails point away from the Sun and underlies the idea of a “solar sail”.

Worked example

At the Earth’s orbit the solar intensity is I ≈ 1370 W/m². What pressure does sunlight exert on a black (fully absorbing) surface? What power does a 1 m² panel of 20 % efficiency give at I = 1000 W/m²?

1. Pressure: p = I/c = 1370/(3·10⁸) ≈ 4.6·10⁻⁶ Pa, about 10¹⁰ times less than atmospheric pressure.
2. Panel power: P = η·I·S = 0.2·1000·1 = 200 W.

Check yourself

1. Energy of a photon of frequency 5·10¹⁴ Hz?

2. The work function is 4.5 eV. Threshold wavelength?

3. 6 eV photons fall on a metal with a 4 eV work function. Stopping voltage?

4. The light intensity is doubled at the same frequency. What changes?

5. Momentum of a photon with λ = 663 nm?

6. 2.5 eV photons fall on a metal with a 3 eV work function. Is there a photoelectric effect?

7. Light of intensity 600 W/m² falls normally on a mirror and is fully reflected. Pressure?

8. In which materials is the internal photoelectric effect observed?

Common mistakes

• Thinking brighter light gives electrons more energy (it gives more electrons).
• Mixing electronvolts and joules (1 eV = 1.6·10⁻¹⁹ J).
• Thinking emission occurs below the threshold frequency: it does not.
• Forgetting to convert nm to metres.
• Setting the kinetic energy equal to hν: the work function must be subtracted.